A family uses 8kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate o

Exercises · 5.23

8\ \text{kW}

(a) Direct solar energy is incident on the horizontal surface at an average rate of

200\ \text{W}

per square meter. If

20\%

of this energy can be converted to useful electrical energy, how large an area is needed to supply

8\ \text{kW}

(b) Compare this area to that of the roof of a typical house.

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Concept understanding — Kinetic Energy Flux

Kinetic Energy Flux: From Intuition to Precision

Imagine you're standing by a river. The water is moving — it carries kinetic energy. Now imagine you hold a net across the flow. How much kinetic energy passes through that net every second? That's the idea behind kinetic energy flux: the rate at which kinetic energy is transported across a surface by a moving fluid (or any moving substance).

The Intuition First

Think of a conveyor belt carrying boxes. Each box has mass $m$ and moves at speed $v$ , so its kinetic energy is $\frac{1}{2}mv^2$ . If boxes pass a point at a rate of $n$ boxes per second, the kinetic energy passing that point per second is $n \times \frac{1}{2}mv^2$ .

Now replace boxes with a continuous fluid — water, air, or even a stream of particles. Instead of counting discrete boxes, we talk about mass flow rate: how much mass crosses a surface per second. If a fluid of density $\rho$ flows with speed $v$ through a cross-sectional area $A$ , the mass flow rate is $\rho v A$ (mass per second). Multiply that by the kinetic energy per unit mass, $\frac{1}{2}v^2$ , and you get the kinetic energy flux:

\text{Kinetic energy flux} = \frac{1}{2} \rho v^3 A

That $v^3$ is striking — it tells you that doubling the speed multiplies the energy flux by eight. That's why wind turbines are so sensitive to wind speed: a small increase in wind speed dramatically increases the power available.

The Precise Statement

In physics and engineering, flux generally means "flow per unit area per unit time." So kinetic energy flux is often defined as the rate of kinetic energy transfer per unit area:

\text{Kinetic energy flux density} = \frac{1}{2} \rho v^3

This is a vector quantity — it points in the direction of flow. If you want the total power (energy per second) crossing a surface of area $A$ oriented perpendicular to the flow, you multiply:

P = \frac{1}{2} \rho A v^3

Watch out

This formula assumes the velocity is uniform across the area and perpendicular to it. If the flow is at an angle, you use the component of velocity normal to the surface: $v_\perp = v \cos\theta$ , giving $P = \frac{1}{2} \rho A (v \cos\theta)^3$ .

Where Does It Come From?

Start with a small parcel of fluid of mass $dm$ moving at speed $v$ . Its kinetic energy is $dK = \frac{1}{2} v^2 dm$ .

In a time $dt$ , the fluid moves a distance $v\,dt$ . The volume that crosses a surface of area $A$ in that time is $A v\, dt$ . The mass in that volume is $dm = \rho A v\, dt$ .

So the kinetic energy crossing in time $dt$ is:

dK = \frac{1}{2} v^2 (\rho A v\, dt) = \frac{1}{2} \rho A v^3 dt

Dividing by $dt$ gives the power (energy flux):

P = \frac{dK}{dt} = \frac{1}{2} \rho A v^3

Why It Matters

Wind energy: The power available in wind is $\frac{1}{2} \rho A v^3$ . This is the fundamental limit for wind turbines.
Fluid dynamics: In pipe flow, kinetic energy flux helps calculate energy losses and pump requirements. …

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The key idea is to use the kinetic energy flux concept — power per unit area — to find the required collection area. For part (a), the useful power per square meter is $20\%$ of $200\ \text{W/m}^2 = 40\ \text{W/m}^2$ , so the area needed is $8000\ \text{W} \div 40\ \text{W/m}^2 = 200\ \text{m}^2$ . For part (b), this is much larger than a typical house roof (say $100\ \text{m}^2$ ), so the answer is that the required area is about twice the roof area.

Why this works: the idea of energy flux

When sunlight falls on a surface, we talk about power per unit area — that’s the energy flux. Think of it like rain: if rain falls at 200 drops per second on each square meter, and you can catch 20% of those drops, then each square meter gives you 40 useful drops per second. To get 8000 drops per second, you need enough square meters.

Here, the “drops” are watts of power. The solar flux is $200\ \text{W/m}^2$ , but only $20\%$ becomes useful electricity. So the useful power per square meter is:

\text{Useful flux} = 0.20 \times 200\ \text{W/m}^2 = 40\ \text{W/m}^2

That’s the rate at which one square meter can supply electrical power.

Step-by-step solution

1. Write down what’s given and what’s needed.

Total power required: $P_{\text{need}} = 8\ \text{kW} = 8000\ \text{W}$
Incident solar flux: $I = 200\ \text{W/m}^2$
Conversion efficiency: $\eta = 20\% = 0.20$

We want the area $A$ such that the collected useful power equals $8000\ \text{W}$ .

2. Express the useful power from an area $A$ .

The total incident power on area $A$ is $I \times A$ . Only $\eta$ of that becomes electrical power:

P_{\text{useful}} = \eta \cdot I \cdot A

3. Set this equal to the required power and solve for $A$ .

\eta I A = P_{\text{need}}

A = \frac{P_{\text{need}}}{\eta I}

Plug in numbers:

$A = \frac{8000\ \text{W}}{0.20 \times 200\ \text{W/m}^2} = \frac{8000}{40}\ \text{m}^2 = 200\ \text{m}^2$ …

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Q.A family uses $8\ \text{kW}$ of power.

Concept understanding — Kinetic Energy Flux

Kinetic Energy Flux: From Intuition to Precision

The Intuition First

The Precise Statement

Where Does It Come From?

Why It Matters

Why this works: the idea of energy flux

Step-by-step solution

Concept: Power per Unit Area (Flux) with Conversion Efficiency

Q.A family uses 8 kW8\ \text{kW}8 kW of power.

Why this works: the idea of energy flux

Step-by-step solution

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Q.A family uses $8\ \text{kW}$ of power.

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